How Does Quantum Mechanics Describe Momentum in Three Dimensions?

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Homework Help Overview

This discussion revolves around the application of quantum mechanics in describing momentum in three dimensions, particularly focusing on the Fourier transform of wave functions in momentum space. Participants are exploring the mathematical representation of momentum and position operators in three-dimensional space.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to extend the one-dimensional Fourier transform to three dimensions and are questioning how to express the inner products of momentum and position states in this context. There is confusion regarding the terms that arise when considering the components of momentum and position vectors.

Discussion Status

The discussion is active, with participants providing insights into the mathematical framework and expressing uncertainty about specific terms and their significance. Some have suggested using linearity of the inner product to clarify the relationships between different components, while others are seeking a deeper understanding of the derivation of certain results.

Contextual Notes

Participants are navigating the complexities of quantum mechanics notation and the implications of multi-dimensional Fourier transforms. There is an acknowledgment of the need for rigorous definitions and the potential for cross terms in the expressions being discussed.

varunag
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In our Quantum mechanics course we were told that the wave function can be projected in the momentum space by taking the Fourier transform of the projection of the wave function in the coordinate space. This was shown in one-dimension as:
\langle p|\psi \rangle = \int_{-\infty}^{\infty} \langle p|x \rangle \langle x|\psi \rangle dx

But, how do we write if we hve to write in three dimensions?
i.e.
\langle \vec{p}|\psi \rangle = \int_{-\infty}^{\infty} \langle \vec{p}|\vec{r} \rangle \langle \vec{r}|\psi \rangle dx
Here I'm confused as to what is \langle \vec{p}|\vec{r} \rangle ?
If I write \vec{p} = p_x \hat{i} + p_y \hat{j} + p_z \hat{k}
and \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}
then should the terms like, \langle p_y|x \rangle crop up in the final result?
 
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In 3 dimensions, the position operator \hat{x} is conventionally denoted by \hat{r} (since the position vector in 3-space is denoted by \vec{r}). Having said that you can still use \hat{x} with the understanding that \hat{x} has three components and that the integral you have written down has to be carried out over each of the 3 spatial coordinates.

Using \hat{r}, I will write this as

\langle p|\psi\rangle = \int d^{3}\vec{r} \langle p|r\rangle \langle r|\psi\rangle

Now, I can use

\langle r|p \rangle = \frac{1}{(2\pi \hbar)^{3/2}}\exp\left(\frac{i}{\hbar}{\vec{p}\bullet \vec{r}}\right)

But I think what you wanted to ask was the significance of the quantity \langle x|p_{y}\rangle.

EDIT: To do it rigorously, you can use linearity of the inner product:

\langle \alpha_{1} + \alpha_{2} | \beta_{1} \rangle = \langle \alpha_{1} | \beta_{1} \rangle + \langle \alpha_{2} | \beta_{1}\rangle

with suitable number of terms, using the expansion of the position and momentum operators in terms of their x, y, z components and then using the expressions for \langle x|p_{x}\rangle, and so on. You will get the cross term \langle x|p_{y}\rangle in such an expression.
 
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The result you have given is known, but I don't know how the result came. i.e. how do we write:
\langle \vec{p}|\vec{r} \rangle = \langle p_x|x \rangle \langle p_y|y \rangle \langle p_z|z \rangle?
 
varunag said:
The result you have given is known, but I don't know how the result came. i.e. how do we write:
\langle \vec{p}|\vec{r} \rangle = \langle p_x|x \rangle \langle p_y|y \rangle \langle p_z|z \rangle?

This follows from the generalization of the Fourier Transform in 3 dimensions. Specifically,

\psi(\vec{r}) = \frac{1}{(2\pi\hbar)^{3/2}} \int d^{3}\vec{p} \exp\left(-\frac{i}{\hbar}\vec{p}\bullet\vec{r}\right)\phi(\vec{p})

\phi(\vec{p}) = \frac{1}{(2\pi\hbar)^{3/2}} \int d^{3}\vec{r} \exp\left(\frac{i}{\hbar}\vec{p}\bullet\vec{r}\right)\psi(\vec{r})

You can write this in Dirac notation easily.

Note: The result is consistent with the fact that [x_{l},p_{m}] = i\hbar \delta_{l,m}.

EDIT: This can be derived by inserting complete sets of states, i.e. writing \langle \vec{p}|\vec{r}\rangle as an integral (multiple integral) over basis states corresponding to the different x, y, z components of p and r.
 

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